Question

After 76.0 min, 31.0% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?

?= min

Answer 1

for a first order reaction,

the half life,t = (ln2)/k.

In order to find half life, we need to find the rate constant, k the equation for a first oreder reaction,

ln[A] = ln[A]o -kt

at time t= 76 min, [A] = (100-31)%[A]o = 69% [A]o or 0.69[A]o

Subs these value into the equation to find the k's value.

ln0.69[A]o = ln[A]o - k (76)

ln0.69[A]o - ln[A]o = - k(76)

ln (0.69[A]o) / [A]o = -k (76)

k = 4.88 * 10^-3

After getting the value of k then use the half life equation for first order to find the half life.

half life,t = (ln2)/k.

             = ln2 / 4.88 * 10^-3

         t = 142 min